Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $z = \dfrac{10k - 10}{k^2 - 5k} \times \dfrac{k^2 - 3k - 10}{k + 2} $
Explanation: First factor the quadratic. $z = \dfrac{10k - 10}{k^2 - 5k} \times \dfrac{(k - 5)(k + 2)}{k + 2} $ Then factor out any other terms. $z = \dfrac{10(k - 1)}{k(k - 5)} \times \dfrac{(k - 5)(k + 2)}{k + 2} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ 10(k - 1) \times (k - 5)(k + 2) } { k(k - 5) \times (k + 2) } $ $z = \dfrac{ 10(k - 1)(k - 5)(k + 2)}{ k(k - 5)(k + 2)} $ Notice that $(k + 2)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 10(k - 1)\cancel{(k - 5)}(k + 2)}{ k\cancel{(k - 5)}(k + 2)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $z = \dfrac{ 10(k - 1)\cancel{(k - 5)}\cancel{(k + 2)}}{ k\cancel{(k - 5)}\cancel{(k + 2)}} $ We are dividing by $k + 2$ , so $k + 2 \neq 0$ Therefore, $k \neq -2$ $z = \dfrac{10(k - 1)}{k} ; \space k \neq 5 ; \space k \neq -2 $